Question: What's the first wrong statement in the proof below that $ \triangle BCE \cong \triangle BCA$ $ \; ?$ $ \overline{BC} $ is parallel to $ \overline{DF} $. This diagram is not drawn to scale. $A$ $B$ $C$ $D$ $E$ $F$ Givens $ \overline{BC} \cong \overline{BD}$ $, \ $ $ \angle ABC \cong \angle DBE$ $, \ $ $ \angle BAC \cong \angle BED$ $, \ $ $ \angle ABC \cong \angle CFE$ $, \ $ $ \overline{AB} \cong \overline{EF}$ $, \ $ and $\ $ $ \angle BAC \cong \angle CEF$ Proof $ \triangle BDE \cong \triangle BCA$ because AAS $ \overline{BE} \cong \overline{AB}$ because corresponding parts of congruent triangles are congruent $ \angle CBE \cong \angle BED$ because alternate interior angles are equal $ \triangle BDE \cong \triangle BCE$ because ASA $ \triangle FCE \cong \triangle BCA$ because ASA $ \triangle BCE \cong \triangle BCA$ because SSS
Answer: Try going through the proof yourself: write down the givens, and then see if they justify the next step for the reason given. Then do the same thing for the next step, and the next, until you run into something that you can't justify, or you finish the proof. $ \triangle BCE \cong \triangle BDE$ is the first wrong statement.